Conditions for solvability

Here, \(a_1/a_2=21/3=7\) and \(b_1/b_2=28/4=7\), whereas \(c_1/c_2=84/15=28/5\). Thus, \(a_1/a_2=b_1/b_2\ne c_1/c_2\), which represents inconsistent equations whose lines are parallel. Therefore, the pair has no solution. Exam tip: When the first two coefficient ratios are equal but the constant-term ratio is different, the pair has no solution.

For solvability by ratio comparison, check whether \(\frac{17}{8}=\frac{10}{5}\) and whether the corresponding constants also keep the same ratio. Compute:
\[\frac{10}{5}=2,\quad \frac{17}{8}=\frac{17}{8}\neq 2.\]
Since the coefficient ratios already do not match, the system is inconsistent and has no solution.

A pair of linear equations is dependent when both equations represent exactly the same line. Such a pair is consistent because it has solutions, but it does not have just one solution; every point on the common line satisfies both equations. To identify this situation, compare all corresponding coefficients and constants after multiplying or dividing an equation by a non-zero number.

Here, multiplying the second equation by \(12\) gives \(12(2x+3y=14)\), or \(24x+36y=168\). This is exactly the first equation. Therefore, the two equations are the same equation and their graphs coincide. They have infinitely many common solutions, so the pair is consistent and dependent. Choice A is correct; “independent” would mean that the lines meet at only one point.

Compare the corresponding coefficients: \(\frac{22}{2}=11\) and \(\frac{33}{3}=11\), but \(\frac{99}{11}=9\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), which is the condition for an inconsistent pair. The two lines are distinct and parallel, so the equations have no solution. Exam tip: compare \(a_1/a_2\), \(b_1/b_2\), and \(c_1/c_2\); the pair is dependent only when all three ratios are equal.

For the given equations, \(a_1/a_2=16/8=2\) and \(b_1/b_2=7/4\). Since \(2\ne 7/4\), the two lines intersect at one point, so the pair has a unique solution. Therefore, it is consistent and independent. A consistent dependent pair would require \(a_1/a_2=b_1/b_2=c_1/c_2\). Exam tip: First compare the ratios of the coefficients of \(x\) and \(y\).

The second equation is (3) times the first. Therefore, both conditions give the same information and have infinitely many solutions.

The ratios of the coefficients are \(\frac{8}{16}=\frac{3}{6}=\frac{1}{2}\), but the ratio of the constant terms is \(\frac{280}{575}=\frac{56}{115}\neq\frac{1}{2}\). Thus, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), so the two lines are parallel and have no common solution; the system is inconsistent. Option B would be correct only if all three ratios were equal, giving infinitely many solutions. Exam tip: Compare the three coefficient and constant ratios directly to identify the type of system quickly.

For the given equations, a₁/a₂ = 6/3 = 2 and b₁/b₂ = 5/(-2) = -5/2. Since a₁/a₂ ≠ b₁/b₂, the two lines intersect at exactly one point, so the pair has a unique solution. Infinitely many solutions would require a₁/a₂ = b₁/b₂ = c₁/c₂, which is not true here. Exam tip: first compare the ratios of the coefficients of x and y.

The ratios of the corresponding coefficients and constants are \(\frac{9}{27}=\frac{16}{48}=\frac{74}{222}=\frac{1}{3}\). Hence, all three ratios are equal. When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the two linear equations represent the same line and have infinitely many solutions. Exam tip: Always compare the ratio of the constant terms as well; checking only the coefficients of x and y is not sufficient.

Here, \(a_1=14, b_1=21, c_1=98\) and \(a_2=2, b_2=3, c_2=17\). We get \(\frac{a_1}{a_2}=\frac{14}{2}=7\) and \(\frac{b_1}{b_2}=\frac{21}{3}=7\), but \(\frac{c_1}{c_2}=\frac{98}{17}\) is not 7. Hence, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), so the two lines are parallel and the pair has no solution. Option B is incorrect because \(\frac{98}{17}\ne7\). Exam tip: compare all three ratios; equality of only the first two ratios means no solution.

Here, (10/20) \ne (11/21), since 10/20 = 1/2 while 11/21 has a different value. Therefore, the two lines intersect at exactly one point, so the pair has a unique solution. Option A would apply when the first two ratios are equal but the constant-term ratio is different; option C requires all three ratios to be equal. Exam tip: for two linear equations, a₁/a₂ \ne b₁/b₂ always indicates a unique solution.

For a pair of linear equations to be consistent and dependent, the corresponding coefficients and constant terms must be in the same ratio. Here, the coefficients in the second equation are three times those in the first: 21=3×7 and 39=3×13. Therefore, its constant term must also be three times 67, so s=3×67=201. Hence, option C is correct. Exam tip: First identify the common multiplier between the corresponding x- and y-coefficients, then apply it to the constant term.

The first two ratios are equal. For inconsistency, the constant ratio must be different so (r \ne 74).

For two linear equations to have no solution, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{8}{16}=\frac{1}{2}\). Therefore, \(\frac{7}{a}=\frac{1}{2}\), which gives \(a=14\). Also, \(\frac{34}{79}\ne\frac{1}{2}\), so the two equations represent distinct parallel lines and have no solution. Exam tip: first equate the ratios of the coefficients of the variables, then verify that the constants have a different ratio.

For two linear equations to have infinitely many solutions, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) must hold. Here, \(\frac{10}{2}=\frac{110}{22}=5\), so \(\frac{k}{9}=5\), which gives \(k=45\). Therefore, option C is correct. Exam tip: compare the constant-term ratio first, then equate the corresponding coefficient ratio to it.

For two linear equations to have no solution, the ratios of the coefficients of the variables must be equal, while the ratio of the constant terms must be different. Thus, \(\frac{l}{18}=\frac{15}{45}=\frac{1}{3}\), which gives \(l=6\). Also, \(\frac{60}{181}\neq\frac{1}{3}\), since \(180\neq181\); therefore, the two lines are distinct and parallel. If 5 or 7 were used, the coefficient ratios would not be equal, resulting in a unique solution. Exam tip: For no solution, remember \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\).

The second equation is exactly twice the first: multiplying 16x+11y=73 by 2 gives 32x+22y=146. Hence, both equations represent the same line and the pair has infinitely many common solutions. Equivalently, \(16/32=11/22=73/146=1/2\). Therefore, option B is correct; equal ratios indicate coincident lines, not distinct parallel lines. Exam tip: if \(a_1/a_2=b_1/b_2=c_1/c_2\), the two lines coincide and the pair has infinitely many solutions.

For the two equations, a₁/a₂ = 20/4 = 5 and b₁/b₂ = (-15)/(-3) = 5, but c₁/c₂ = 85/18, which is not 5. Thus, the two lines have equal slopes but different intercepts, so they are distinct parallel lines and never intersect. Therefore, the pair has no solution. Infinitely many solutions require all three ratios to be equal, while a unique solution requires the first two ratios to be unequal. Exam tip: If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the pair is inconsistent and has no solution.

For the two equations, the determinant of the coefficients is \(a_1b_2-a_2b_1=23\times5-11\times10=5\). Since it is non-zero, the two lines intersect at exactly one point, so the pair has one unique solution. Option B would apply when the two lines coincide, while option A would apply to distinct parallel lines. Exam tip: if \(\frac{a_1}{a_2}\ne\frac{b_1}{b_2}\), a pair of linear equations has a unique solution.

The first two ratios are equal. For inconsistency, the constant ratio must be different so (t \ne 93).

For two linear equations to have infinitely many solutions, the ratios of the corresponding coefficients and constants must be equal: 5/20 = a/28 = 45/180. Since 5/20 and 45/180 are both 1/4, we get a/28 = 1/4, so a = 7. Therefore, option C is correct. Exam tip: For infinitely many solutions, all three corresponding ratios must be equal; equality of only two ratios is not sufficient.

For two linear equations to have no solution, the condition is \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\). Here, \(\frac{b}{16}=\frac{9}{24}=\frac{3}{8}\), giving \(b=6\). Also, \(\frac{36}{97}\ne\frac{3}{8}\), so the two lines are distinct and parallel. Exam tip: For ‘no solution’, first equate the ratios of the coefficients of the variables, then verify that the constants’ ratio is different.

Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have a unique solution when \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\). Here, we require \(\frac{13}{6} \ne \frac{p}{5}\). Thus, \(65 \ne 6p\), or \(p \ne \frac{65}{6}\). Therefore, option A is correct. In option B, the two ratios become equal, so it does not give a unique solution. Exam tip: For a unique solution, compare the ratios of the coefficients of the two variables and ensure they are unequal.

Here, \(\frac{a_1}{a_2}=\frac{14}{21}=\frac{2}{3}\) and \(\frac{b_1}{b_2}=\frac{-8}{-12}=\frac{2}{3}\), but \(\frac{c_1}{c_2}=\frac{25}{40}=\frac{5}{8}\). Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines are parallel and distinct, so the pair has no solution. In an exam, apply this ratio condition directly.

Here, \(\frac{a_1}{a_2}=\frac{9}{27}=\frac{1}{3}\) and \(\frac{b_1}{b_2}=\frac{4}{12}=\frac{1}{3}\), but \(\frac{c_1}{c_2}=\frac{360}{1090}\neq\frac{1}{3}\). Thus, the two lines are distinct and parallel, so they have no solution and the system is inconsistent. Option B would be correct only if all three ratios were equal, giving infinitely many solutions. Exam tip: When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\), the system is inconsistent.